Derivatives of cosec x, sec x and cot x: deriving them with the chain and quotient rules, the chain-rule forms, and using them for tangents, normals and stationary points (new in the 2024 Mathematics Advanced syllabus)

Write $\operatorname{cosec} x = (\sin x)^{-1}$ and use the chain rule:

Markers look for the chain rule (or quotient rule on $\frac{1}{\sin x}$) set out correctly, and a final line that visibly splits $\frac{\cos x}{\sin^2 x}$ into $\operatorname{cosec} x \cot x$. In a "show that" question the splitting step is the mark, so do not skip it.

$\frac{dy}{dx} = \sec^2 x - \operatorname{cosec}^2 x$. Setting this to zero gives $\sec^2 x = \operatorname{cosec}^2 x$, so $\frac{1}{\cos^2 x} = \frac{1}{\sin^2 x}$, which means $\tan^2 x = 1$. In $0 < x < \frac{\pi}{2}$, $\tan x > 0$, so $\tan x = 1$ and $x = \frac{\pi}{4}$.

Then $y = \tan \frac{\pi}{4} + \cot \frac{\pi}{4} = 1 + 1 = 2$.

Nature: $\frac{d^2y}{dx^2} = 2 \sec^2 x \tan x + 2 \operatorname{cosec}^2 x \cot x$. At $x = \frac{\pi}{4}$ this is $2(2)(1) + 2(2)(1) = 8 > 0$, so $\left( \frac{\pi}{4}, 2 \right)$ is a minimum turning point.

Markers award one mark for the derivative, one for solving to $x = \frac{\pi}{4}$ (rejecting $\tan x = -1$ with a reason), one for $y = 2$ and one for a valid test of nature.

Differentiate term by term with the standard results. $\frac{d}{dx}(\sec x) = \sec x \tan x$ and $\frac{d}{dx}(\cot x) = -\operatorname{cosec}^2 x$, so

Marker's note: one mark for each correct term. The usual slip is dropping the negative sign on the $\cot x$ term.

A constant multiple stays in front. Using $\frac{d}{dx}(\operatorname{cosec} x) = -\operatorname{cosec} x \cot x$,

Marker's note: one mark for the derivative of $\operatorname{cosec} x$, one for the constant and the sign.

Chain rule with the inner function $u = 5x - 1$. Since $\frac{d}{du}(\cot u) = -\operatorname{cosec}^2 u$ and $\frac{du}{dx} = 5$,

Marker's note: one mark for $-\operatorname{cosec}^2(5x - 1)$, one for multiplying by the inner derivative $5$.

Product rule with $u = x^2$ and $v = \sec x$. Here $u' = 2x$ and $v' = \sec x \tan x$, so

Factor out the common $x \sec x$.

Marker's note: one mark for setting up the product rule, one for the correct derivative of $\sec x$, one for the factored answer.

Differentiate with the chain rule. With $u = \sec x$, $\frac{d}{dx} \ln u = \frac{u'}{u}$:

On $-\frac{\pi}{2} < x < \frac{\pi}{2}$ we have $\sec x > 0$, so the logarithm is defined.

Hence integrate. $\ln(\sec x)$ is a primitive of $\tan x$, so

Marker's note: one mark for the derivative, one for recognising the primitive, one for $\ln 2$ (using $\sec \frac{\pi}{3} = 2$).

Find the point. $y = \sec \frac{\pi}{4} = \frac{1}{\cos (\pi/4)} = \sqrt{2}$, so the point is $\left( \frac{\pi}{4}, \sqrt{2} \right)$.

Find the gradient of the tangent. $\frac{dy}{dx} = \sec x \tan x$, so at $x = \frac{\pi}{4}$,

The normal is perpendicular, so its gradient is $-\frac{1}{\sqrt{2}}$.

Point-gradient form.

Marker's note: one mark for the point, one for the tangent gradient $\sqrt{2}$, one for the normal gradient, one for the equation. Using the tangent gradient in the final line is the classic error.

(a) Differentiate. $f^\prime(x) = \sec x \tan x + 2 \sin x$. Write $\sec x \tan x = \frac{\sin x}{\cos^2 x} = \sin x \sec^2 x$, so

(b) Stationary points. $\sec^2 x + 2 > 0$ for every $x$ in the domain, so $f^\prime(x) = 0$ only when $\sin x = 0$, that is $x = 0$. Then $f(0) = 1 - 2 = -1$.

Nature: for $x$ slightly less than $0$, $\sin x < 0$ so $f^\prime(x) < 0$; for $x$ slightly more than $0$, $f^\prime(x) > 0$. The gradient changes from negative to positive, so $(0, -1)$ is a minimum turning point.

(c) Range. As $x \to \pm \frac{\pi}{2}$, $\sec x \to \infty$ while $2 \cos x \to 0$, so $f(x) \to \infty$. With a single minimum value of $-1$, the range is $f(x) \geq -1$.

Marker's note: one mark for $f^\prime$, one for rewriting it in the required form, one for $x = 0$ and $f(0) = -1$, one for justifying the minimum, one for the range.

Differentiate. $\frac{dy}{dx} = -\operatorname{cosec}^2 x$.

Use the Pythagorean identity $1 + \cot^2 x = \operatorname{cosec}^2 x$ (divide $\sin^2 x + \cos^2 x = 1$ by $\sin^2 x$):

Marker's note: one mark for the derivative, one for the identity, one for substituting $y = \cot x$ to finish.